About stabilizer in group action Let $X$ be a finite set and $x$ is an element of $X$. Let $G_x$, the stabilizer subgroup, be the subset of $S_X$ consisting of permutations that fix $x$.
The question is 

Is stabilizer always a normal subgroup?

I know that $G_x=\{g\in G\mid gx=x, x \in X\}$ and I have proved that stabilizer is a subgroup of $G$. But I got stuck at the point proving stabilizer is always normal.
I have found that 'if the group action is transitive, then the stabilizer is normal'.
So, I came to a conclusion that stabilizers are not always normal. But I cannot understand this. Is there some good explanation on this (stabilizer is not always normal) with examples?
Thank you.
 A: Take the action of $S_3$ on itself by conjugation and $x=(1\ 2)$. Then the stabilizer of $x$ is $\{e, x\}$ and this is not a normal subgroup of $S_3$.
A: If this where true all subgroups would be normal. Take a subgroup $H$ of $G$. Consider the left regular action of $G$ on the set of left cosets of $G$ by $H$. The stabilizer of $H$ is clearly $H$. Using the above we conclude $H$ is normal.
A: Consider the symmtric group on $X$. Conjugation here corresponds to 'renaming' the elements of $X$. Suppose as usual $X=\{1,2,\dots,n\}$. So for example conjugating the cycle $(1234)$ by the transposition $(24)$ gives the cycle $(1432)$, because we've renamed '4' to be '2' and vice versa.
So a normal subgroup of $S_X$ is one which has the same elements in it when you do any 'renaming' of the symbols $1,\dots,n$ throughout its elements. You can see then that a stabilizer of a point here is the last thing we'd expect to be normal. The subgroup $G_4$ consisting of elements that fix $4$, say, contains some elements that fix $2$ and some that don't, so it won't consist of the same elements if '4' and '2' swap roles. (It will conjugate to $G_2$, of course.)
